It is known that there is no (nontrivial) partition of $\Bbb{R}$ into a countable number of closed set. But is there a partition of $\Bbb{R}$ into sets, each one homeomorphic to the cantor ternary set?
2 Answers
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Let $f$ be the x-coordinate of Hilbert's space-filling curve,
whose graph is shown here:
(source: osu.edu)
Then the sets $\{f^{-1}(t)\}_{t\in [0,1]}$
form a partition of the interval [0,1] into Cantor sets.
An easy variation of the above construction produces a partition of the reals (take the point-preimages of $F$, where $F:\mathbb R\to \mathbb R$ is the periodic extension given by $F(t):=\lfloor t\rfloor+f(t-\lfloor t\rfloor)$).
answered Sep 11, 2013 at 11:37
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1$\begingroup$ Not clear what kind of a function do you mean by "like this". Since the concept is about uncountably uncountable sets more exact argument needed. $\endgroup$– user37834Commented Sep 11, 2013 at 11:45
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3$\begingroup$ Silvi: I agree with you, and I've edited my answer. It's now completely constructive. $\endgroup$ Commented Sep 11, 2013 at 12:06
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2$\begingroup$ Well, it's "constructive" in some sense of the word constructive which is not "constructive mathematics". But it is a beautiful answer! $\endgroup$ Commented Sep 11, 2013 at 12:56
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$\begingroup$ another picture of the graph commons.wikimedia.org/wiki/File:Hilbertfunction_y.gif $\endgroup$ Commented Jan 5, 2023 at 8:49
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$\begingroup$ There is no proof (or even a claim) in the link of the statement that point-preimages of components of (the/a) Hilbert space-filling curve are homeomorphic to the Cantor set. How do you prove it? $\endgroup$ Commented Jan 5, 2023 at 14:26
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Various questions about partitioning a topological space $X$ into homeomorphic copies of a topological space $Y$ are discussed by Paul Bankston and Richard J. McGovern, Topological partitions, General Topology and its Applications 10 (1979), 215–229 (pdf). In particular, their Theorem 1.14 says that $\mathbb R$ can be partitioned into Cantor sets. The proof consists of first partitioning $\mathbb R$ into countably many Cantor sets and one set which is isomorphic to the space $\mathbb P$ of irrational numbers, and then using the fact that $\mathbb P$ can be partitioned into $\mathfrak c$ Cantor sets, i.e., $\mathbb N^\mathbb N$ can be partitioned into $\mathfrak c$ copies of $\{0,1\}^\mathbb N$.
From the fact that $\mathbb R$ can be partitioned into Cantor sets, it easily follows that every nonempty perfect subset of $\mathbb R$ can be partitioned into $\mathfrak c$ Cantor sets. Bankston and McGovern prove more general theorems under special set-theoretic assumptions. Namely, they use Martin's axiom to prove that every complete separable metric space with no isolated points can be partitioned into Cantor sets, and they use the continuum hypothesis to prove the same for complete metric spaces of cardinality $\mathfrak c$ with no isolated points.
In my answer to the Math.SE question Partitioning a metric space into Cantor sets, I proved in ZFC that a nonempty Polish space with no isolated points can be partitioned into $\mathfrak c$ Cantor sets.
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1$\begingroup$ Hi bof, just to let you know, I recently proved a few theorems that improve on the work of Bankston and McGovern that you mention in your second paragraph. It turns out their theorems are provable in ZFC (you just have to work a bit harder without MA or CH), and in fact can be strengthened as well. The paper is here: sciencedirect.com/science/article/abs/pii/S0166864122002826. The main results are: (1) If a metric space has size $\leq \mathfrak{c}^{+\omega}$, then it can be partitioned into copies of the Cantor space if and only if it can be covered with copies of the Cantor space. $\endgroup$ Commented Jan 5, 2023 at 12:48
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1$\begingroup$ (2) the same is true for first countable spaces of size $\mathfrak{c}$. (3) the same is true for completely metrizable spaces of any size. Furthermore, for (1) and (2) there are examples that show these cardinality bounds are necessary and sharp. That is, there is a metrizable space of size $\mathfrak{c}^{+\omega+1}$ that can be covered with copies of the Cantor space, but not partitioned into copies of the Cantor space; and there is a first countable space of size $\mathfrak{c}^+$ with this property as well. $\endgroup$ Commented Jan 5, 2023 at 12:50